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This is proven in Rademacher and Toeplitz (1957).

(b-c)/a = (sin[1/2(B-C)])/(cos(1/2A)) (1) (c-a)/b = (sin[1/2(C-A)])/(cos(1/2B)) (2) (a-b)/c = (sin[1/2(A-B)])/(cos(1/2C)). (3)

Let a triangle have side lengths a, b, and c with opposite angles A, B, and C. Then (b+c)/a = (cos[1/2(B-C)])/(sin(1/2A)) (1) (c+a)/b = (cos[1/2(C-A)])/(sin(1/2B)) (2) ...

A perfect cubic polynomial can be factored into a linear and a quadratic term, x^3+y^3 = (x+y)(x^2-xy+y^2) (1) x^3-y^3 = (x-y)(x^2+xy+y^2). (2)

sum_(k=-infty)^infty(a; m-k)(b; n-k)(a+b+k; k)=(a+n; m)(b+m; n).

The problem of finding all independent irreducible algebraic relations among any finite set of quantics.

By the definition of the trigonometric functions, cos0 = 1 (1) cot0 = infty (2) csc0 = infty (3) sec0 = 1 (4) sin0 = 0 (5) tan0 = 0. (6)

By the definition of the functions of trigonometry, the sine of pi is equal to the y-coordinate of the point with polar coordinates (r,theta)=(1,pi), giving sinpi=0. ...

cos(pi/(10)) = 1/4sqrt(10+2sqrt(5)) (1) cos((3pi)/(10)) = 1/4sqrt(10-2sqrt(5)) (2) cot(pi/(10)) = sqrt(5+2sqrt(5)) (3) cot((3pi)/(10)) = sqrt(5-2sqrt(5)) (4) csc(pi/(10)) = ...

cos(pi/(12)) = 1/4(sqrt(6)+sqrt(2)) (1) cos((5pi)/(12)) = 1/4(sqrt(6)-sqrt(2)) (2) cot(pi/(12)) = 2+sqrt(3) (3) cot((5pi)/(12)) = 2-sqrt(3) (4) csc(pi/(12)) = sqrt(6)+sqrt(2) ...